Answer
$b = -4, b = 7$
Work Step by Step
Use either the square root method, the quadratic formula or factoring. $$
\begin{aligned}
& b^2-3 b=28 \\
& b^2-3 b+\left(\frac{3}{2}\right)^2=28+\left(\frac{3}{2}\right)^2 \\
& \left(b-\frac{3}{2}\right)^2=\frac{4}{4} \cdot 28+\frac{9}{4} \\
& \left(b-\frac{3}{2}\right)^2=\frac{121}{4} \\
& \left(b-\frac{3}{2}\right)= \pm \sqrt{\frac{121}{4}}\\
&\left(b-\frac{3}{2}\right) = \pm \frac{11}{2} \\
& b=\frac{3}{2} \pm \frac{11}{2}.
\end{aligned}
$$ This gives: $$
\begin{aligned}
b & =\frac{3+11}{2} \\
& =7\\
b & =\frac{3-11}{2} \\
& =-4.
\end{aligned}
$$ Define the following version of the function and plot it. We see that the zeros are where they should be.
$$
f(b) = b^2-3 b-28.
$$
